Question 2.49: Use nodal analysis to determine the current flowing through ......
Use nodal analysis to determine the current flowing through the various branches in the circuit shown in Fig. 2.158. All resistances shown are in Ohms.
We have shown the current directions in the various branches of the circuit and will apply KCL of node B, node C, and node D, respectively. Let V_B,\ V_C,\ V_D be the voltages at these nodes.
We have
or, 2 \mathrm{~V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{c}}-10=0 (i)
Then
\mathrm{I}_3=\mathrm{I}_4+\mathrm{I}_5 \text { at node } \mathrm{C}Putting values
\frac{V_B-V_C}{10}=\frac{V_C-12}{20}+\frac{V_C-V_D}{20}or, 2 V_B-4 V_C+V_D+12=0 (ii)
Again
\begin{gathered} \mathrm{I}_5+\mathrm{I}_7=\mathrm{I}_6 \\ \frac{\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{D}}}{20}+0.5=\frac{\mathrm{V}_{\mathrm{D}}}{20} \end{gathered}or, V_C-V_D+10=V_D
or, \mathrm{V}_{\mathrm{c}}-2 \mathrm{~V}_{\mathrm{D}}+10=0 (iii)
Solving Eqs. (i), (ii) and (iii), we get
\begin{aligned} & \mathrm{V}_{\mathrm{B}}=10.4 \mathrm{~V}, \quad \mathrm{~V}_{\mathrm{C}}=10.8 \mathrm{~V}, \quad \mathrm{~V}_{\mathrm{D}}=10.4 \mathrm{~V} \\ & \mathrm{I}_2=\frac{\mathrm{V}_{\mathrm{B}}}{10}=\frac{10.4}{10}=1.04 \mathrm{~A} \\ & \mathrm{I}_3=\frac{\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{C}}}{10}=\frac{10.4-10.8}{10}=-\frac{0.4}{10}=-0.04 \mathrm{~A} \end{aligned}(direction of I_2 is opposite to that shown)
\begin{aligned} & I_5=\frac{V_C-V_D}{20}=\frac{10.8-10.4}{20}=\frac{0.4}{20}=0.02 \mathrm{~A} \\ & I_6=\frac{V_D}{20}=\frac{10.4}{20}=0.52 \mathrm{~A} \end{aligned}Again \mathrm{I}_6=\mathrm{I}_7+\mathrm{I}_5=0.5+0.02=0.52 \mathrm{~A}
again \mathrm{I}_2=\mathrm{I}_1+\mathrm{I}_3=1.0+0.04=1.04 A
\begin{aligned} & \mathrm{I}_4+\mathrm{I}_5=\mathrm{I}_3 \\ & \begin{aligned} \mathrm{I}_4 & =\mathrm{I}_3-\mathrm{I}_5 \\ & =0.04-0.02 \\ & =0.02 \mathrm{~A} \end{aligned} \end{aligned}